Is there a simple (ish) approximation for the expected number of steps a random walk is within a set of bounds over a given time period? - in particular if i presume log normal and constant vol.

If i want to do it with something like local/stochastic/both vol, then is there a way to do it without resorting to MC/diffusion?

  • $\begingroup$ Can you remind us what is the payoff of a corridor? $\endgroup$ – Gordon Jun 28 '16 at 15:14
  • $\begingroup$ The way I interpret this is the Stopping Time $\tau$ until $x(\tau) \geq B1 ∨ x(\tau) \leq B2$ $\endgroup$ – noob2 Jun 28 '16 at 15:29
  • $\begingroup$ Sorry, should have been clearer - it's the total number of days it is within the corridor, it does not knock out. $\endgroup$ – will Jun 28 '16 at 15:50

This looks to me like a range accrual. Let $t_1, \ldots, t_n$, where $0 < t_1 < \cdots < t_n$ be business days that are being considered. We compute \begin{align*} E\left(\sum_{i=1}^n \pmb{1}_{b_1 < S_{t_i} < b_2} \right) &=\sum_{i=1}^n E\left(\pmb{1}_{b_1 < S_{t_i} < b_2} \right)\\ &=\sum_{i=1}^n \left[E\big(\pmb{1}_{S_{t_i} > b_1}\big) -E\big(\pmb{1}_{S_{t_i} \ge b_2} \big) \right]\\ &=\sum_{i=1}^n \left[\Phi\big(d_2^i(b_1)\big) - \Phi\big(d_2^i(b_2)\big)\right], \end{align*} where, for positive constant $K$, \begin{align*} d_2^i(K) = \frac{\ln\frac{S_0}{K} + \big(r-\frac{1}{2}\sigma^2 \big) t_i}{\sigma \sqrt{t_i}}. \end{align*} For general stochastic volatility, there is no simple method. Monte Carlo may be needed.

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  • $\begingroup$ Okay, I thought was afraid MC might have been the only way. $\endgroup$ – will Jun 28 '16 at 20:45
  • $\begingroup$ For the constant vol version though, this is presuming that each possible accrual day is a completely unrelated event to the say before, which is completely not the case, it's path dependant. $\endgroup$ – will Jun 28 '16 at 20:47
  • $\begingroup$ @will: Yes, unless you have a dterministic volatility function. $\endgroup$ – Gordon Jun 28 '16 at 20:47
  • $\begingroup$ presumably only a subset of deterministic vol functions are solvable in and simple (ish) way. $\endgroup$ – will Jun 28 '16 at 20:48
  • 2
    $\begingroup$ They are indeed correlated, but the number of steps is a sum, and the expectation is additive, unless you can define your understanding of expected number of steps more specifically, my answer is what I can interpret. $\endgroup$ – Gordon Jun 28 '16 at 20:59

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